Video: Finding the First Derivative of an Exponential Function

Use the integral test to determine whether the series βˆ‘_(𝑛 = 1)^(∞) (1/𝑛) converges or diverges.

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Video Transcript

Use the integral test to determine whether the series, which is the sum from 𝑛 equals one to ∞ of one over 𝑛, converges or diverges.

Let’s start by recalling the integral test. The integral test tells us that if 𝑓 is a continuous, positive, decreasing function on the interval between π‘˜ and ∞ and that 𝑓 of 𝑛 is equal to π‘Ž 𝑛. Then, if the integral from π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is convergent, so is the sum from 𝑛 equals π‘˜ to ∞ of π‘Ž 𝑛. And secondly, if the integral from π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is divergent, so is the sum from 𝑛 equals π‘˜ to ∞ of π‘Ž 𝑛. Now, in our case, our series is the sum from 𝑛 equals one to ∞ of one over 𝑛. Therefore, π‘Ž 𝑛 is equal to one over 𝑛. Using the fact that 𝑓 of 𝑛 is equal to π‘Ž 𝑛, we can say that 𝑓 of π‘₯ is equal to one over π‘₯.

Now, we need to check that 𝑓 of π‘₯ is a continuous, positive, and decreasing function on the interval between π‘˜ and ∞. Noting that since our sum goes from 𝑛 equals one, π‘˜ must be equal to one. Let’s start by checking the continuity of our function over this interval. The only discontinuity of 𝑓 of π‘₯ occurs when π‘₯ is equal to zero. However, zero is not in our interval between one and ∞. Therefore, this function must be continuous between one and ∞. On this interval, π‘₯ is always positive. Therefore, one over π‘₯ must also be positive. And so, we’ve satisfied that condition too. Now, if π‘₯ starts at one and gets larger and larger and larger, then one over π‘₯ will get smaller and smaller and smaller. Therefore, we can say that this is a decreasing function on our interval.

Now that we’ve satisfied these three conditions, we’re able to use the integral test. We need to work out whether the integral from one to ∞ of one over π‘₯ with respect to π‘₯ is convergent or divergent. Now, we know that the differential of the natural logarithm of π‘₯ with respect to π‘₯ is equal to one over π‘₯. Therefore, the natural logarithm of π‘₯ is the antiderivative of one over π‘₯. Hence, we have that our integral is equal to the natural logarithm of π‘₯ between one and ∞. When using the bound of ∞, we need to take the limit as π‘₯ tends to ∞ of the natural logarithm of π‘₯. Then, for our lower bound of one, we simply subtract the natural logarithm of one.

Now, let’s evaluate this limit, the natural logarithm as an increasing function. Therefore, as π‘₯ gets larger, the natural logarithm of π‘₯ also gets larger. Hence, we can say that this limit is also equal to ∞. We also have that the natural logarithm of one is equal to zero. However, regardless of what constant this term is, this will not affect our answer of ∞. This tells us that our integral must be divergent. Hence, by the integral test, we can say that the sum from 𝑛 equals one to ∞ of one over 𝑛 is divergent.

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